# Commutative Algebra/Basics on prime and maximal ideals and local rings

## Prime ideals

Definition 12.1:

Let ${\displaystyle R}$ be a ring. A prime ideal ${\displaystyle p}$ is an ideal of ${\displaystyle R}$ such that whenever ${\displaystyle ab\in p}$, either ${\displaystyle a\in p}$ or ${\displaystyle b\in p}$.

Lemma 12.2:

Let ${\displaystyle R}$ be a ring and ${\displaystyle I\leq R}$ an ideal. ${\displaystyle I}$ is prime if and only if ${\displaystyle R/I}$ is an integral domain.

Proof:

${\displaystyle I}$ prime is equivalent to ${\displaystyle ab\in I\Rightarrow a\in I\vee b\in I}$. This is equivalent to

${\displaystyle ab+I=0+I\Rightarrow a+I=0+I\vee b+I=0+I}$.${\displaystyle \Box }$

Theorem 12.3:

Let ${\displaystyle S\subseteq R}$ be multiplicatively closed. Then there exists a prime ideal not intersecting ${\displaystyle S}$.

Proof:

Order all ideals of ${\displaystyle R}$ not intersecting ${\displaystyle S}$ by set inclusion, and let a chain

${\displaystyle I_{1}\subseteq I_{2}\subseteq \cdots \subseteq I_{k}\subseteq \cdots }$

be given. The ideal

${\displaystyle I:=\bigcup _{k\in \mathbb {N} }I_{k}}$

(this is an ideal, since ${\displaystyle a,b\in I\Rightarrow a\in I_{m},b\in I_{n}\Rightarrow a,b\in I_{\max\{m,n\}}}$, hence ${\displaystyle a+b\in I}$, ${\displaystyle ra\in I}$) is an upper bound of the chain, since ${\displaystyle I}$ cannot intersect ${\displaystyle S}$ for else one of the ${\displaystyle I_{k}}$ would intersect ${\displaystyle S}$. Since the given chain was arbitrary, Zorn's lemma implies the existence of a maximal ideal among all ideals not intersecting ${\displaystyle S}$. This ideal shall be called ${\displaystyle J}$; we prove that it is prime.

Let ${\displaystyle ab\in J}$, and assume for contradiction that ${\displaystyle a\notin J}$ and ${\displaystyle b\notin J}$. Then ${\displaystyle \langle a\rangle +J}$, ${\displaystyle \langle b\rangle +J}$ are strict superideals of ${\displaystyle J}$ and hence intersect ${\displaystyle S}$, that is,

${\displaystyle s=xa+yj}$,
${\displaystyle t=zb+wj'}$,

${\displaystyle s,t\in S}$, ${\displaystyle x,y,z,w\in R}$, ${\displaystyle j,j'\in J}$. Then ${\displaystyle S\ni st=xzab+yzbj+xawj'+ywjj'\in J}$, contradiction.${\displaystyle \Box }$

## Projection to the quotient ring

In this section, we want to fix a notiation. Let ${\displaystyle R}$ be a ring and ${\displaystyle I\leq R}$ an ideal. Then we may form the quotient ring ${\displaystyle R/I}$ consisting of the elements of the form ${\displaystyle r+I}$, ${\displaystyle r\in R}$. Throughout the book, we shall use the following notation for the canonical projection ${\displaystyle r\mapsto r+I}$:

Definition 12.4:

Let ${\displaystyle I\leq R}$ an ideal. The map

${\displaystyle \pi _{I}:R\to R/I,\pi _{I}(r):=r+I}$

is the canonical projection of ${\displaystyle R}$ to ${\displaystyle R/I}$.

## Maximal ideals

Definition 12.5:

Let ${\displaystyle R}$ be a ring. A maximal ideal ${\displaystyle m}$ of ${\displaystyle R}$ is an ideal that is not the whole ring, and there is no proper ideal ${\displaystyle I\leq R}$ such that ${\displaystyle m\subsetneq I}$.

Lemma 12.6:

An ideal ${\displaystyle I\leq R}$ is maximal iff ${\displaystyle R/I}$ is a field.

Proof:

A ring is a field if and only if its only proper ideal is the zero ideal. For, in a field, every nonzero ideal contains ${\displaystyle 1}$, and if ${\displaystyle R}$ is not a field, it contains a non-unit ${\displaystyle a}$, and then ${\displaystyle \langle a\rangle }$ does not contain ${\displaystyle 1}$.

By the correspondence given by the correspondence theorem, ${\displaystyle R/I}$ corresponds to ${\displaystyle R}$, the zero ideal of ${\displaystyle R/I}$ corresponds to ${\displaystyle I}$, and any ideal strictly in between corresponds to an ideal ${\displaystyle K\leq R}$ such that ${\displaystyle I\subsetneq K\subsetneq R}$. Hence, ${\displaystyle R/I}$ is a field if and only if there are no proper ideals strictly containing ${\displaystyle I}$.${\displaystyle \Box }$

Lemma 12.7:

Any maximal ideal is prime.

Proof 1:

If ${\displaystyle R}$ is a ring, ${\displaystyle m\leq R}$ maximal, then ${\displaystyle R/m}$ is a field. Hence ${\displaystyle R/m}$ is an integral domain, hence ${\displaystyle m}$ is prime.${\displaystyle \Box }$

Proof 2:

Let ${\displaystyle m\leq R}$ be maximal. Let ${\displaystyle ab\in m}$. Assume ${\displaystyle a,b\notin m}$. Then ${\displaystyle 1=ra+sn=tb+uk}$ for suitable ${\displaystyle n,k\in m}$, ${\displaystyle r,s,t,u\in R}$. But then ${\displaystyle 1=1^{2}=(ra+sn)(tb+uk)=rtab+stbn+rauk+sunk\in m}$.${\displaystyle \Box }$

Theorem 12.8:

Let ${\displaystyle R}$ be a ring and ${\displaystyle I\leq R}$ an ideal not equal to all of ${\displaystyle R}$. Then there exists a maximal ${\displaystyle m\leq R}$ with ${\displaystyle I\subseteq m}$.

Proof:

We order the set of all ideals ${\displaystyle J}$ such that ${\displaystyle I\subseteq J}$ and ${\displaystyle J\neq R}$ by inclusion. Let

${\displaystyle J_{1}\subseteq J_{2}\subseteq \cdots \subseteq J_{k}\subseteq \cdots }$

be a chain of those ideals. Then set

${\displaystyle J:=\bigcup _{k\in \mathbb {N} }J_{k}}$.

Clearly, all ${\displaystyle J_{k}}$ are contained within ${\displaystyle J}$. Since ${\displaystyle I\subseteq J_{1}}$, ${\displaystyle I\subseteq J}$. Further, assume ${\displaystyle 1\in J}$. Then ${\displaystyle 1\in J_{m}}$ for some ${\displaystyle m}$, contradiction. Hence, ${\displaystyle J}$ is a proper ideal such that ${\displaystyle I\subseteq J}$, and hence an upper bound for the given chain. Since the given chain was arbitrary, we may apply Zorn's lemma to obtain the existence of a maximal element with respect to inclusion. This ideal must then be maximal, for any proper superideal also contains ${\displaystyle I}$.${\displaystyle \Box }$

Lemma 12.9:

Let ${\displaystyle R}$ be a ring, ${\displaystyle I\leq R}$. Then via ${\displaystyle \pi _{I}}$, maximal ideals of ${\displaystyle R}$ containing ${\displaystyle I}$ correspond to maximal ideals of ${\displaystyle R/I}$.

Proof: From the correspondence theorem.${\displaystyle \Box }$

## Local rings

Definition 12.10:

A local ring is a ring that has exactly one maximal ideal.

Theorem 12.11 (characterisation of local rings):

Let ${\displaystyle R}$ be a ring. The following are equivalent:

1. ${\displaystyle R}$ is a local ring.
2. If ${\displaystyle a+b}$ is a unit, then either ${\displaystyle a}$ or ${\displaystyle b}$ is a unit, where ${\displaystyle a,b\in R}$ arbitrary.
3. The set of all non-units forms a maximal ideal.
4. If ${\displaystyle r_{1},\ldots ,r_{n}\in R}$ where ${\displaystyle r_{1}+\cdots +r_{n}}$ is a unit, then one of the ${\displaystyle r_{j}}$ is a unit.
5. If ${\displaystyle r\in R}$ is arbitrary, either ${\displaystyle r}$ or ${\displaystyle 1-r}$ is a unit.

Proof:

1. ${\displaystyle \Rightarrow }$ 2.: Assume ${\displaystyle a}$ and ${\displaystyle b}$ are both non-units. Then ${\displaystyle \langle a\rangle }$ and ${\displaystyle \langle b\rangle }$ are proper ideals of ${\displaystyle R}$ and hence they are contained in some maximal ideal of ${\displaystyle R}$ by theorem 12.7. But there is only one maximal ideal ${\displaystyle m}$ of ${\displaystyle R}$, and hence ${\displaystyle a,b\in m}$, thus ${\displaystyle a+b\in m}$. Maximal ideals can not contain units.

2. ${\displaystyle \Rightarrow }$ 3.: The sum of two non-units is a non-unit, and if ${\displaystyle a}$ is a non-unit and ${\displaystyle r\in R}$, ${\displaystyle ra}$ is a non-unit (for if ${\displaystyle sra=1}$, ${\displaystyle sr}$ is an inverse of ${\displaystyle a}$). Hence, all non-units form an ideal. Any proper ideal of ${\displaystyle R}$ contains only non-units, hence this ideal is maximal.

3. ${\displaystyle \Rightarrow }$ 4.: Assume the ${\displaystyle r_{j}}$ are all non-units. Since the non-units form an ideal, ${\displaystyle r_{1}+\cdots +r_{n}}$ is contained in that ideal of non-units, contradiction.

4. ${\displaystyle \Rightarrow }$ 5.: Assume ${\displaystyle r}$, ${\displaystyle 1-r}$ are non-units. Then ${\displaystyle 1=r+(1-r)}$ is a non-unit, contradiction.

5. ${\displaystyle \Rightarrow }$ 1.: Let ${\displaystyle m,n\leq R}$ two distinct maximal ideals. Then ${\displaystyle m+n=R}$, hence ${\displaystyle 1=s+t}$, ${\displaystyle s\in m}$, ${\displaystyle t\in n}$, that is, ${\displaystyle s=1-t}$. ${\displaystyle t}$ is not a unit, so ${\displaystyle 1-t=s}$ is, contradiction.${\displaystyle \Box }$

## Localisation at prime ideals

In chapter 9, we had seen how to localise a ring at a multiplicatively closed subset ${\displaystyle S}$. An important special case is ${\displaystyle S=R\setminus p}$, where ${\displaystyle p}$ is a prime ideal.

Lemma 12.12:

Let ${\displaystyle p\leq R}$ be a prime ideal of a ring. Then ${\displaystyle S:=R\setminus p\subseteq R}$ is multiplicatively closed.

Proof: Let ${\displaystyle a,b\notin p}$. Then ${\displaystyle ab}$ can't be in ${\displaystyle p}$, hence ${\displaystyle ab\in S}$.${\displaystyle \Box }$

Definition 12.13:

Let ${\displaystyle p\leq R}$ be a prime ideal of a ring. Set ${\displaystyle S:=R\setminus p}$. Then

${\displaystyle R_{p}:=S^{-1}R}$

is called the localisation of ${\displaystyle R}$ at ${\displaystyle p}$.

Theorem 12.14:

Let ${\displaystyle R}$ be a ring, ${\displaystyle p\leq R}$ be prime. ${\displaystyle R_{p}}$ is a local ring.

Proof:

Set ${\displaystyle S:=R\setminus p}$, then ${\displaystyle R_{p}=S^{-1}R}$. Set

${\displaystyle m:=\{r/s|r\in p,s\in S\}\subseteq S^{-1}R}$.

All elements of ${\displaystyle m}$ are non-units, and all elements of ${\displaystyle R_{p}\setminus m}$ are of the form ${\displaystyle r'/t}$, ${\displaystyle r'\notin p}$, ${\displaystyle t\in S}$ and thus are units. Further, ${\displaystyle m}$ is an ideal since ${\displaystyle p}$ is and by definition of addition and multiplication in ${\displaystyle S^{-1}R}$ and since ${\displaystyle S}$ is multiplicatively closed. Hence ${\displaystyle R_{p}}$ is a local ring.${\displaystyle \Box }$

This finally explains why we speak of localisation.